Solving Momentum Problem: 4000N/m Spring Compression

  • Thread starter Thread starter mikep
  • Start date Start date
  • Tags Tags
    Momentum
Click For Summary
SUMMARY

The discussion focuses on solving a physics problem involving a 10kg cart with a 4000N/m spring. Initially compressed by 0.50m, the spring's potential energy is calculated as 500J. Upon release, the cart collides with another identical cart, and momentum conservation is applied to determine the velocity of the combined carts. Finally, using the kinetic energy of the combined carts, the compression of the second spring is calculated to be 0.25m.

PREREQUISITES
  • Understanding of potential energy and kinetic energy concepts
  • Knowledge of conservation of momentum principles
  • Familiarity with elastic potential energy calculations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn about elastic potential energy and its applications in spring systems
  • Explore advanced momentum conservation scenarios in collisions
  • Investigate real-world applications of springs in engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and engineers involved in dynamics and energy systems will benefit from this discussion.

mikep
Messages
43
Reaction score
0
can someone please help me with this problem?
A 10kg cart with low friction wheels has a 4000N/m
spring on one end. The cart is pressed against the
wall so that the spring is compressed 0.50m. The cart
is released and rolls on the surface only to hit
another identical cart. The two carts lock together
and run into a wall on the other side of the room
where a 4000N/m spring on the second cart compresses
until the carts stop. How far does the second spring
compress as the carts stop?
I think i need to break it up into three parts. For
the first one i need to find the force of the s[ring
then the collision of hte two carts and then the
compression of the spring. So for the first part i
did:
PEs = (1/2) k(x^2) = (1/2)(4000N/m)(0.50^2) = 1000J
is that right for the first part?
 
Last edited:
Physics news on Phys.org
Check your math. I think you forgot to divide by 2. And the unit for energy is the joule. I think your strategy is essentially correct. Now that you know the initial potential energy of the cart, you know its final kinetic energy as it is released (from conservation of energy). Therefore, you know its velocity when it hit the second cart. From conservation of momentum, you can calculate the velocity of the two combined carts after the collision. That will give you their initial kinetic energy when they hit the spring. Since this kinetic energy is converted to elastic potential energy, you can calculate the displacement of the spring.
 
oh you're right its 500J. ok so i use that and i did 500 = (1/2)m(v^2) v = 10m/s then i did the conservation of momentum. (10)(10) + (10)(0) = (10 + 10) V V = 5m/s (1/2)m(V^2) = (1/2)k(x^2) x = 0.25 as the displacement. does this look right?
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
17
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K